Question

Common roots of the equations $\theta$ and $2n\pi + \frac{7\pi}{6}$ are.

• A. $2\sin^{2}x + \sin^{2}2x = 2$
• B. $\sin 2x + \cos 2x = \tan x$
• C. $\sin^{2}2x = 2\cos^{2}x$
• D. None of these

Answer 1

Answer: (B)

$\sin 2x + \cos 2x = \tan x$

Answer 2

$12\pi$ ......(i)

and $\cot 3x$ .....(ii)

Solving (i), $\frac{\pi}{3}$

⇒$\cos(4x + 3)$ ⇒$\frac{\pi}{2}$

$\Rightarrow$Common roots are $\pi$

Solving (ii), $2\sin 3\theta$

$\frac{2\pi}{3}4\cos 3\theta$

$\Rightarrow$ $\frac{\pi}{3}$ $f(x) = \sin^{4}x + \cos^{4}x$ $(\sin^{2}x + \cos^{2}x)^{2} - 2\sin^{2}x\cos^{2}x$

Trick : For $1 - \frac{4\sin^{2}x\cos^{2}x}{2} = 1 - \frac{\sin^{2}2x}{2}$, option (1) gives $1 - \frac{1}{4}(2\sin^{2}2x) = 1 - \left( \frac{1 - \cos x}{4} \right)$ which satisfies the equation (i) but does not satisfy the (ii). Now option (2) gives $= \frac{3}{4} + \frac{1}{4}\cos 4x$ which satisfies both the equations.